A stone is projected from the ground with velocity `50(m)/(s)` at an angle of `30^@`. It crosses a wall after 3 sec. How far beyond the wall the stone will strike the ground `(g=10(m)/(sec^2)`?

Options

A. 90.2 m
B. 89.6 m
C. 86.6 m
D. 70.2 m

Explanation

To solve the problem step by step, we will follow the physics of projectile motion.

### Step 1: Determine the initial velocity components
The stone is projected with an initial velocity \( u = 50 \, \text{m/s} \) at an angle \( \theta = 30^\circ \).

The horizontal and vertical components of the initial velocity can be calculated as follows:
- \( u_x = u \cos \theta = 50 \cos 30^\circ = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \, \text{m/s} \)
- \( u_y = u \sin \theta = 50 \sin 30^\circ = 50 \times \frac{1}{2} = 25 \, \text{m/s} \)

### Step 2: Calculate the total time of flight
The total time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting the values:
\[ T = \frac{2 \times 50 \times \frac{1}{2}}{10} = \frac{50}{10} = 5 \, \text{s} \]

### Step 3: Determine the time after crossing the wall
The stone crosses the wall after \( t_1 = 3 \, \text{s} \). The remaining time \( t_2 \) for which the stone is in the air after crossing the wall is:
\[ t_2 = T - t_1 = 5 \, \text{s} - 3 \, \text{s} = 2 \, \text{s} \]

### Step 4: Calculate the horizontal distance traveled after crossing the wall
The horizontal distance \( d \) traveled after crossing the wall can be calculated using the horizontal component of the velocity and the time \( t_2 \):
\[ d = u_x \cdot t_2 \]
Substituting the values:
\[ d = (25\sqrt{3}) \cdot 2 = 50\sqrt{3} \, \text{m} \]

Calculating \( 50\sqrt{3} \):
\[ d \approx 50 \times 1.732 \approx 86.6 \, \text{m} \]

### Conclusion
The stone will strike the ground approximately \( 86.6 \, \text{m} \) beyond the wall.

---